Considering (F, G, H) to be coordinates in 3-D space...

The first equation is of a sphere with centre at (0, 0, 0) and radius r, while the second equation is of a plane whose normal has direction ratios (2, 3, 6).

The distance of the plane from the origin is+/- 7r/sqrt(2^{2} + 3^{2} + 6^{2}),

which simplifies to+/- r, so the plane can only touch the sphere at the diametrically opposite points A (2r/7, 3r/7, 6r/7) and B (-2r/7, -3r/7, -6r/7).

Substitution into the second given equation, 1/F + 1/G + 1/H = 1, then shows that only A gives a possible solution, viz r = 7/2 + 7/3 + 7/6 = 7, from which the only possible triple (F, G, H) is (2, 3, 6).